৪৯তম বিসিএস ⎯ প্রাণ রসায়ন [৬০১]

Answer: C
Explanation: Enzymes exhibit remarkable substrate specificity. This is due to the unique three-dimensional structure of their active site, which is precisely shaped to bind to a specific substrate, much like a lock and key. Enzymes are catalysts and are not consumed; they do not change the reaction's equilibrium, only the rate at which it is reached; and they are large.

Explanation: Oxidoreductases (EC 1) are the class of enzymes that catalyze oxidation-reduction (redox) reactions. A common example is lactate dehydrogenase, which catalyzes the interconversion of pyruvate and lactate with the concomitant interconversion of NADH and NADH H+

Explanation: Hydrolases (EC 3) are enzymes that catalyze the cleavage of a chemical bond through the addition of a water molecule (a process called hydrolysis). Proteases like pepsin and chymotrypsin are classic examples of hydrolases.

Answer: C
Explanation: Ligases (EC 6) catalyze the joining (ligation) of two molecules, a reaction that is coupled with the hydrolysis of a high-energy phosphate bond in ATP or another nucleotide. This energy input is required to form the new chemical bond, such as the phosphodiester bond formed by DNA ligase.

Answer: C 
Explanation: The induced-fit model refines the older "lock-and-key" theory. It posits that the active site is flexible. The initial binding of the substrate induces a subtle change in the enzyme's three-dimensional structure, optimizing the alignment of catalytic groups and strengthening the binding interaction, which facilitates catalysis.

Explanation: The International Unit (IU) is a standard measure of enzyme activity. It is precisely defined as the amount of enzyme that catalyzes the formation of one micromole of product per minute under specified assay conditions (e.g., pH, temperature, substrate concentration).

Answer: B
Explanation: The turnover number (kcat​) is a measure of an enzyme's maximum catalytic efficiency. It tells us how many reactions a single active site can catalyze per unit of time when it is fully saturated with substrate. It is calculated as kcat​=Vmax​/[Et​], where [Et​] is the total enzyme concentration.

Explanation: An enzyme's activity is highly dependent on pH. Deviations from the optimum pH alter the protonation state of ionizable amino acid side chains (like Asp, Glu, His, Lys). This can disrupt the precise ionic and hydrogen bonds maintaining the enzyme's crucial 3D structure (leading to denaturation) and can also affect the charge of residues directly involved in substrate binding and catalysis at the active site.

Explanation: According to collision theory, increasing the temperature increases the kinetic energy of molecules. This results in more frequent collisions between the enzyme and substrate and ensures that more of these collisions have sufficient energy to overcome the activation energy barrier. Beyond the optimal temperature, the enzyme rapidly denatures, causing a sharp drop in activity.

Answer: B Explanation: When the substrate is in excess (saturating), the reaction rate is limited only by the amount of enzyme available to process it. Therefore, if you double the concentration of the enzyme, you double the number of active sites available, and the overall reaction rate (V0​) will also double. This demonstrates a direct, linear relationship.

Answer: C
Explanation: The Michaelis-Menten equation is a cornerstone of enzyme kinetics. It mathematically models how the initial rate of reaction (V0​) changes in response to varying concentrations of the substrate ([S]), for a fixed amount of enzyme.

Answer: C
Explanation: By definition, Km​ is the substrate concentration required for the enzyme to achieve half of its maximum velocity (Vmax​/2). It is a characteristic property of an enzyme for a particular substrate and is an indicator of the affinity between them.

Answer: D
Explanation: Vmax​ is the theoretical maximum rate when all enzyme active sites are saturated with substrate. This rate is limited by the speed of catalysis. Therefore, the more enzyme molecules you have, the higher the maximum possible rate. This relationship is expressed as Vmax = kcat​[Et​].

Answer: C Explanation: The specificity constant (kcat​/Km​) combines the enzyme's catalytic rate (kcat​) and its affinity for the substrate (inversely related to Km​). It describes how efficiently an enzyme converts substrate into product at low substrate concentrations, making it the most comprehensive measure of catalytic performance.

Answer: C
Explanation: A competitive inhibitor "competes" with the substrate for the same active site because it typically has a chemical structure similar to the substrate. This binding is reversible, and its effect can be overcome by increasing the substrate concentration, which will outcompete the inhibitor for access to the active site.

Answer: B
Explanation: A non-competitive inhibitor binds to an allosteric (non-active) site on the enzyme, regardless of whether the substrate is bound. This binding alters the enzyme's conformation, reducing its catalytic efficiency without affecting its ability to bind the substrate. As a result, Vmax​ is decreased, but the affinity for the substrate, and thus Km​, remains unchanged.

Answer: C
Explanation: Uncompetitive inhibition is a unique case where the inhibitor can bind only to the pre-formed enzyme-substrate (ES) complex, forming an inactive ESI complex. This action effectively removes the ES complex from the reaction, leading to a decrease in both Vmax​ and Km​.

Answer: D
Explanation: Irreversible inhibitors, also known as inactivators or suicide substrates, bind very tightly to an enzyme, often by forming a covalent bond with a key amino acid residue in the active site. This permanently inactivates the enzyme molecule.

Answer: B) 2 ATP
Explanation: Glycolysis produces 4 ATP, but 2 ATP are consumed during the preparatory phase, resulting in a net gain of 2 ATP.

Answer: A) Hexokinase
Explanation: Hexokinase catalyzes the first step of glycolysis, converting glucose to glucose-6-phosphate using 1 ATP.

Answer: C) AMP
Explanation: AMP signals low energy status and activates PFK-1, promoting glycolysis. ATP and citrate inhibit it.

Answer: C) Pyruvate
Explanation: In aerobic conditions, glucose is broken down into two molecules of pyruvate, which then enter the TCA cycle.

Answer: C) Glyceraldehyde-3-phosphate → 1,3-bisphosphoglycerate
Explanation: This reaction, catalyzed by G3P dehydrogenase, produces NADH.

Answer: B) Substrate-level phosphorylation
Explanation: Pyruvate kinase converts phosphoenolpyruvate (PEP) to pyruvate, producing ATP directly.

Answer: C) ATP
Explanation: ATP acts as a negative feedback inhibitor, indicating high energy availability, thus slowing glycolysis.

Answer: B) 2
Explanation: Each glyceraldehyde-3-phosphate produces 1 NADH, and since 2 G3P are formed per glucose, the total is 2 NADH.

Answer: B) Citrate
Explanation: Acetyl-CoA combines with oxaloacetate to form citrate, catalyzed by citrate synthase.

Answer: B) 2
Explanation: Each acetyl-CoA entering the cycle results in the release of two CO₂ molecules.

Answer: B) NADH and CO₂
Explanation: This enzyme catalyzes oxidative decarboxylation of isocitrate to α-ketoglutarate, releasing NADH and CO₂.

Answer: C) 10
Explanation:

3 NADH = 7.5 ATP
1 FADH₂ = 1.5 ATP
1 GTP = 1 ATP
Total = 10 ATP

answer: B) Succinate → Fumarate
Explanation: Succinate dehydrogenase catalyzes this reaction, reducing FAD to FADH₂.

Answer: C) NADH
Explanation: High levels of NADH signal sufficient energy, inhibiting key enzymes like isocitrate dehydrogenase.

Answer: C) Pyruvate dehydrogenase
Explanation: The pyruvate dehydrogenase complex (PDC) catalyzes this key irreversible step.

Answer: C) NADPH and Ribose-5-phosphate
Explanation: NADPH is used for biosynthesis, and Ribose-5-phosphate is essential for nucleotide synthesis.

Answer: B) Glucose-6-phosphate dehydrogenase
Explanation: This enzyme converts glucose-6-phosphate to 6-phosphoglucono-δ-lactone, producing NADPH.

Answer: B) Hemolytic anemia
Explanation: G6PD deficiency reduces NADPH, impairing glutathione regeneration, leading to oxidative damage to RBCs.

Answer: B) Interconversion of sugars
Explanation: This phase generates various sugars like ribose-5-phosphate and fructose-6-phosphate.

Answer: B) Liver and kidney
Explanation: These organs maintain blood glucose levels during fasting by synthesizing glucose.

Answer: B) Pyruvate carboxylase and PEP carboxykinase
Explanation: Pyruvate → Oxaloacetate (via pyruvate carboxylase) → PEP (via PEP carboxykinase).

Answer: C) Pyruvate carboxylase
Explanation: This enzyme converts pyruvate to oxaloacetate using ATP and biotin.

Answer: D) Acetyl-CoA
Explanation: Acetyl-CoA cannot be converted back to glucose because the pyruvate dehydrogenase reaction is irreversible.

Answer: C) Activate gluconeogenesis
Explanation: High ATP indicates sufficient energy, shifting metabolism toward glucose production rather than breakdown.

Answer: B) AMP
Explanation: AMP signals low energy status and inhibits gluconeogenesis, favoring glycolysis instead.

Explanation: (b) Alpha-ketoglutarate is the primary amino group acceptor in most transamination reactions, forming glutamate. Glutamate (a) is often the donor in subsequent reactions. Oxaloacetate (c) and pyruvate (d) can also act as acceptors in specific transaminase reactions, but alpha-ketoglutarate is the most universal.

Explanation: (c) Glutamate dehydrogenase catalyzes the oxidative deamination of glutamate, using NAD⁺ (or NADP⁺) as a cofactor, which results in the production of alpha-ketoglutarate and free ammonium ion (NH₄⁺). This is a key step in channeling amino nitrogen into the urea cycle.

Explanation: (b) Lysine and leucine are the only two amino acids that are exclusively ketogenic, meaning their carbon skeletons are degraded only to acetyl-CoA or acetoacetate, not to glucose precursors. Phenylalanine (c) and isoleucine (d) are both glucogenic and ketogenic.

Explanation: (b) The urea cycle is a hepatic (liver) process. Its first two steps occur in the mitochondrial matrix, and the subsequent three steps occur in the cytosol.

Explanation: (b) Transamination is a reversible reaction that involves the transfer of an α-amino group from an amino acid to an α-ketoacid, resulting in a new amino acid and a new ketoacid. This is the primary mechanism for redistributing nitrogen.

Explanation: Carbamoyl phosphate is synthesized in the mitochondrial matrix by the enzyme carbamoyl phosphate synthetase I (CPS I), which uses CO₂, NH₄⁺ (ammonium ion), and 2 ATP molecules.